What volume of carbon dioxide can be obtained by the interaction of 20 g of limestone containing

What volume of carbon dioxide can be obtained by the interaction of 20 g of limestone containing 97% of impurities with hydrochloric acid?

1.Let’s find the mass of pure CaCO3 limestone.

100% – 97% = 3%.

20 g – 100%,

X – 3%,

X = (3% × 20 g): 100% = 0.6 g.

2.Let’s find the amount of substance CaCO

n = m: M.

M (CaCO3) = 100 g / mol.

n = 0.6 g: 100 g / mol = 0.006 mol.

Let’s find the quantitative ratios of substances.

CaCO3 + 2HCl → CaCl2 + H2O + CO2 ↑.

For 1 mol of CaCO3, there is 1 mol of CO2.

Substances are in quantitative ratios 1: 1.

The amount of substance will be equal.

n (CO2) = n (CaCO3) = 0.006 mol.

3.Let’s find the volume of CO2.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.006 mol × 22.4 L / mol = 0.1344 L.

Answer: 0.1344 l.